Acausality in solving time-domain inhomogeneous differential equations with Fourier transforms? I was always wondering about the acausal nature of solutions obtained by Fourier transforms in the case of inhomogeneous equations. The solution usually revolves around the integration of the transform of the inhomegeneous term - and that transform necessarily depends on all future values of that term - so is it really breaking causality?
Example: Analysis of an integrator circuit
Consider a resistor $R$ and capacitor $C$ connected in series to each other, and an external voltage $V(t)$ is applied to the circuit. To find the voltage drop across the capacitor at any moment, we must solve the equation $V(t) = \dot{Q}R + \frac{1}{C}Q$. We may transform the equation to the frequency domain and obtain that
$Q_\omega = \frac{CV_\omega}{1+i\omega RC}$
so the final solution for the voltage across the capacitor would be (using the unitary FT convention):
$V_C=\frac{Q(t)}{C}=\frac{1}{\sqrt{2\pi}}\int{\frac{V_\omega d\omega}{1+i\omega RC}}e^{i\omega t}$
But expanding the term $V_\omega$ clearly shows it involves the integration of $V(t)$ from the dawn till the end of time. This would imply that the solution depends on future values of the input function. Is this really acausal?
Note: Of course, one may take the limit, either $\omega << RC$ or $\omega >> RC $, of the solution and execute the inverse transform analytically and obtain a solution in terms of either $V(t)$ or it's time integral until time $t$, thus removing the problem of causality. But I'm talking about this as a general difficulty, and it's implications on other problems as well.
 A: I) It is not surprising that a solution that uses time-frequency Fourier transformation can superficially look acausal (without actually being acausal), because the Fourier transform 
$$V_{\omega}~ :=~ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\! dt~V(t)e^{-i\omega t} $$
depends by definition on all times $t$ from the far past to the far future.
II) However, the differential equation 
$$V(t) ~=~ R\dot{Q}(t) + \frac{Q(t)}{C}$$ 
can also be solved in a manifestly causal form
$$ Q(t)~=~Q(t_i)\exp\left( \frac{t_i-t}{RC}\right) + \int_{t_i}^{t} \! dt^{\prime}~ \frac{V(t^{\prime})}{R}\exp\left( \frac{t^{\prime}-t}{RC}\right), $$
where we integrate time $t^{\prime}$ from the initial time $t_i$ to now $t$, i.e. only in the past.
A: Causality in the Fourier domain is manifest in the behavior of the transfer function as a complex function of frequency, i.e. the location of poles etc. Your example has a pole at +i/RC, so one may deform the integration countour into the lower complex plane for positive tau and show that it vanishes.
A: 
This would imply that the solution depends on future values of the
  input function.

This isn't true, your reasoning here is faulty.
First, an LTI system is causal if:
$h(t) = 0, t < 0$
where $h(t)$ is the impulse response.
Consider the RC network given.  The impulse response is:
$h(t) = \frac{e^{\frac{-t}{RC}}}{RC}u(t)$
where $u(t)$ is the unit step so this system is causal.
The fact that the transform involves integration over all time does not imply acausality.  For example, consider the identity system where $v_o(t) = v_i(t)$ which is clearly causal.
Now,
$v_o(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\! ~V_i(\omega)e^{j\omega t}d\omega = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\! ~V_o(\omega)e^{j\omega t}d\omega$
Since $V_o$ involves "integration of $v_o(t)$ from the dawn till the end of time", by your reasoning, the value of $v_o(t)$ depends on future values of itself!
